Electronic – Transformer Magnetization Flux

transformer

As far as I understand, an ideal transformer wouldn't show any net flux inside its core, as any such net flux would induce electric fields and in turn currents in the windings such that it will disappear. So an ideal transformer would show no (net) flux at all, correct? So it would kind of act like a control loop trying to keep the flux at 0.

A real transformer requires energy in order to magnetize its core. This energy is provided to the transformer in the form of the magnetizing current \$ I_\mu \$ and a corresponding magnetizing flux \$ \phi_m \$.

It was introduced to me as $$ \phi_m = \phi_1 – \phi_2 $$

with \$\phi_1 \$ being the flux put into the system by the primary winding, and \$\phi_2 \$ being the induced flux coming back from the secondary winding. So what's left after taking this difference is the magnetization flux.

We were shown this visualization in class:
enter image description here

It suggests this flux flows in the whole ferrite core. My internal model was that the energy provided by the additional current \$ I_\mu \$ would be "absorbed" by the core in order to magnetize itself and the flux wouldn't make it to the secondary winding. If it does flow throughout the whole core, and also through the secondary winding, wouldn't it play into the "control loop" described above and vanish as soon as it arises?

What am I missing? Thank you for your time in advance!

Best Answer

My internal model was that the energy provided by the additional current Iμ would be "absorbed" by the core in order to magnetize itself and the flux wouldn't make it to the secondary winding.

Magnetization flux couples both primary and secondary. It induces voltage in the secondary as per the turns ratio. If the secondary is open circuit then that should be easy to see.

The problem most engineers have is realizing that when the secondary is attached to a load, the current in the secondary would "appear" to produce another flux that would "appear" to alter the core flux and screw around with the magnetization.

It doesn't because as soon as the secondary current forms, an extra primary current forms in opposition to the secondary current and the two individual fluxes caused by the secondary current cancel.

What is left is (still) the same magnetization flux and we still get a voltage transformation as per the turns ratio.

So an ideal transformer would show no (net) flux at all, correct?

This is called an ideal power converter or impedance transformer and, as much as EEs like to break down things into smaller manageable lumps, I don't think "an ideal transformer" brings anything to the party when trying to understand non-ideal transformers.

A real transformer requires energy in order to magnetize its core.

I think this misses the point a little. A real transformer has a secondary winding and, if this secondary winding isn't connected to a load, it might just as well be not there at all. So, the "real-transformer" is really just an inductor when it comes to what current it draws from an AC supply in order to produce core magnetism. Nothing more complicated than that.

I mean... do we say that an inductor requires energy to magnetize its core? No we don't; we say that current flows due to the applied voltage and the inductive reactance and that current (along with the number of turns) produces a H field that magnetizes the core. We don't need to think in transformer terms when thinking about core flux. And we don't need to think about energy when defining core flux; current and turns is sufficient.